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I attempted the problem below using mesh analysis, but I'm struggling to figure out how to solve and find the impedance Z. I feel like I'm forgetting something very basic, but I can't figure out what for the life of me. Any help or links to similar questions already answered are greatly appreciated! Thank you!

Edit: My work is attached below.

Circuit Image

Work

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  • \$\begingroup\$ You've selected your answer. But for comparison I get the solution to the mesh as: solve(Eq(simplify(solve([Eq(vs-r1*ia-z3*(ia-ib),0),Eq(-z3*(ib-ia)-r2*ib,0)],[ia,ib])[ia]).subs({ r1:1, r2:10, vs:complex(*(solve([Eq(atan2(y,x),15/180*pi),Eq(sqrt(x**2+y**2),30)],[x,y])[0])) }),5),z3)[0]. Sympy reports: 7.64350779569818 + 5.26451039777369*I But this is also \$9.28107113 \:\angle 34.5573534^\circ\$. \$\endgroup\$ Commented Sep 1, 2023 at 9:47
  • \$\begingroup\$ @periblepsis By hand I obtained the value $$7.64350779569818 + 5.26451039777369$$ without the multiplication by I. How does that relate to the answer $$9.28107113∠34.5573534^∘$$ \$\endgroup\$
    – Farcher
    Commented Sep 1, 2023 at 13:02
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    \$\begingroup\$ @Farcher They are one and the same. The complex number is a vector. It's vector length is the square root of the sum of the squares of the real and imaginary axis values -- pythagorean theorem. The angle is from the atan2() of the same. Just two ways of saying the same thing. One is cartesian, the other polar. \$\endgroup\$ Commented Sep 1, 2023 at 17:11
  • \$\begingroup\$ @periblepsis How stupid of me for not realising what your capital I was. I only thought of current. Many thanks for your answer. \$\endgroup\$
    – Farcher
    Commented Sep 1, 2023 at 18:52

4 Answers 4

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HINT: Given I1, you can find VR1 = I1 * 1. Then you can find VR3 = Vs - Vr1 and thus, IR3 = VR3/10. Finally, I2 = I1 - I3 and V2 = Vs + VR1. Beware that they're all phasors.

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  • \$\begingroup\$ This makes sense, but I'm not understanding your equation for V2. Could you clarify why V2 is not the same as V3 given voltages in parallel are the same? \$\endgroup\$
    – Sam Fuller
    Commented Aug 31, 2023 at 20:59
  • \$\begingroup\$ @SamFuller: it is indeed. you'll be getting the same answer \$\endgroup\$
    – edmz
    Commented Sep 1, 2023 at 20:29
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Hints: -

  • Assume the voltage phase angle is zero degrees for convenience
  • This makes the current phase angle 30° lagging
  • This means that Z is formed from an inductor

What you need to do is figure out the value of inductance that creates this 30° lagging current given that an inductor naturally takes a 90° lagging current. This means that its effect is not over-bearing in the circuit.

You could even convert the voltage source and 1 ohm resistor to a current source as a means to an end.

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Steps for calculating the required impedance: enter image description here

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Well, in general we know that:

$$\underline{\text{V}}=\underline{\text{I}}\cdot\underline{\text{Z}}\tag1$$

When the absolute value and argument of a current, voltage and/or impedance are given, we need to take a look at the following formula's:

  • $$\left|\underline{\text{V}}\right|=\left|\underline{\text{I}}\cdot\underline{\text{Z}}\right|=\left|\underline{\text{I}}\right|\cdot\left|\underline{\text{Z}}\right|\tag2$$
  • $$\arg\left(\underline{\text{V}}\right)=\arg\left(\underline{\text{I}}\cdot\underline{\text{Z}}\right)=\arg\left(\underline{\text{I}}\right)+\arg\left(\underline{\text{Z}}\right)\tag3$$

Where I used easily proviable equalities for the absolute values and arguments of complex numbers.


Now, we apply this to your problem. Let's first take a look at the argument:

$$\arg\left(\underline{\text{V}}_{\space\text{s}}\right)=\arg\left(\underline{\text{I}}_{\space1}\right)+\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\space\Longrightarrow\space15^\circ=0^\circ+\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)\space\Longleftrightarrow\space\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)=15^\circ\tag4$$

The fact that the argument of the impedance is between \$0^\circ\$ and \$90^\circ\$, we can conclude that:

$$\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)>0\space\wedge\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)>0\tag5$$

So, the argument of the impedance is given by:

$$\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)=\arctan\left(\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}\right)=15^\circ\space\Longleftrightarrow\space\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}=2-\sqrt{3}\tag6$$

Using formula \$(2)\$, we can see that:

$$\left|\underline{\text{V}}_{\space\text{s}}\right|=\left|\underline{\text{I}}_{\space1}\right|\cdot\left|\underline{\text{Z}}_{\space\text{i}}\right|\space\Longrightarrow\space30=5\cdot\left|\underline{\text{Z}}_{\space\text{i}}\right|\space\Longleftrightarrow\space\left|\underline{\text{Z}}_{\space\text{i}}\right|=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}=6\space\Omega\tag7$$

We can see that:

$$ \begin{alignat*}{1} \underline{\text{Z}}_{\space\text{i}}&=\text{R}_1+\left(\text{R}_2\space\text{||}\space\underline{\text{Z}}\right)\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\cdot\underline{\text{Z}}}{\displaystyle\text{R}_2+\underline{\text{Z}}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}}\cdot\frac{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}}{\displaystyle\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\left(\Re\left(\underline{\text{Z}}\right)+\Im\left(\underline{\text{Z}}\right)\text{j}\right)\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)-\Im\left(\underline{\text{Z}}\right)\text{j}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\Re^2\left(\underline{\text{Z}}\right)+\text{R}_2\Im^2\left(\underline{\text{Z}}\right)+\text{R}_2^2\Re\left(\underline{\text{Z}}\right)+\text{R}_2^2\Im\left(\underline{\text{Z}}\right)\text{j}}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\\ \\ &=\text{R}_1+\frac{\displaystyle\text{R}_2\Re^2\left(\underline{\text{Z}}\right)+\text{R}_2\Im^2\left(\underline{\text{Z}}\right)+\text{R}_2^2\Re\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}+\frac{\displaystyle\text{R}_2^2\Im\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}\cdot\text{j}\\ \\ &=\underbrace{\text{R}_1+\frac{\displaystyle\text{R}_2\left(\text{R}_2\Re\left(\underline{\text{Z}}\right)\left(1+\Re\left(\underline{\text{Z}}\right)\right)+\Im^2\left(\underline{\text{Z}}\right)\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}}_{\space:=\space\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}+\underbrace{\frac{\displaystyle\text{R}_2^2\Im\left(\underline{\text{Z}}\right)}{\displaystyle\left(\text{R}_2+\Re\left(\underline{\text{Z}}\right)\right)^2+\Im^2\left(\underline{\text{Z}}\right)}}_{\space:=\space\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}\cdot\text{j} \end{alignat*} \tag8 $$

Where \$\displaystyle\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.

So, using \$(5)\$, \$(6)\$, \$(7)\$ and \$(8)\$ we can set-up a system of equations:

$$ \begin{cases} \begin{alignat*}{1} 6&=\sqrt{\Re^2\left(\underline{\text{Z}}_{\space\text{i}}\right)+\Im^2\left(\underline{\text{Z}}_{\space\text{i}}\right)}\\ \\ 2-\sqrt{3}&=\frac{\Im\left(\underline{\text{Z}}_{\space\text{i}}\right)}{\Re\left(\underline{\text{Z}}_{\space\text{i}}\right)}\\ \\ \Re\left(\underline{\text{Z}}_{\space\text{i}}\right)&>0\\ \\ \Im\left(\underline{\text{Z}}_{\space\text{i}}\right)&>0 \end{alignat*} \end{cases}\space\Longleftrightarrow\space\begin{cases} \Re\left(\underline{\text{Z}}_{\space\text{i}}\right)=3\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}\\ \\ \Im\left(\underline{\text{Z}}_{\space\text{i}}\right)=3\sqrt{2-\sqrt{3}} \end{cases}\tag9 $$

And solve for \$\Re\left(\underline{\text{Z}}\right)\$ and \$\Im\left(\underline{\text{Z}}\right)\$:

$$\Re\left(\underline{\text{Z}}\right)\approx7.64351\space\wedge\space\Im\left(\underline{\text{Z}}\right)\approx5.26451\tag{10}$$

The exact values are:

$$\Re\left(\underline{\text{Z}}\right)=\frac{10 \left(36981375 \sqrt{2}+1275 \left(20293 \sqrt{2}+20724\right) \sqrt{3}-10816931\right)}{197063761}\tag{11}$$ $$\Im\left(\underline{\text{Z}}\right)=\frac{150 \left(2103684-450433 \sqrt{2}+574992 \sqrt{3}+1818217 \sqrt{6}\right)}{197063761}\tag{12}$$

I used Mathematica to solve this. The code is provided below:

In[1]:=Clear["Global`*"];
x = 1 + ((10*(a + b*I))/(10 + a + b*I));
\[Alpha] = FullSimplify[ComplexExpand[Re[x]]];
\[Beta] = FullSimplify[ComplexExpand[Im[x]]];
FullSimplify[
 Solve[{6 == Sqrt[\[Alpha]^2 + \[Beta]^2], 
   2 - Sqrt[3] == \[Beta]/\[Alpha], a > 0 && b > 0}, {a, b}, Reals]]

Out[1]={{a -> (10 (-10816931 + 36981375 Sqrt[2] + 
      1275 Sqrt[3] (20724 + 20293 Sqrt[2])))/197063761, 
  b -> (150 (2103684 - 450433 Sqrt[2] + 574992 Sqrt[3] + 
      1818217 Sqrt[6]))/197063761}}

In[2]:=N[%1]

Out[2]={{a -> 7.64351, b -> 5.26451}}
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